Question

The heat of vaporization of 1–pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K•mol. What is the approximate boilingpoint of 1–pentanol?

Answer 1

Answer:

375 K

Explanation:

Using the experssion shown below as:

$\Delta G^0=\Delta H^0_{vap}-T\Delta S^0_{vap}$

At vaporization point, the liquid and the gaseous phase is in the equilibrium.

Thus,

$\Delta G^0=0$

So,

$Delta H^0_{vap}=T\Delta S^0_{vap}$

Given that:

$Delta H^0_{vap}=55.5\ kJ/mol$

Also, 1 kJ = 10³ J

So,

$Delta H^0_{vap}=55500\ J/mol$

$\Delta S^0_{vap}=148\ J/K.mol$

So, temperature is :

$T=\frac{Delta H^0_{vap}}{\Delta S^0_{vap}}$

$T=\frac{55500}{148}$

T= 375 K